Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = 
..............1
put here value in equation 1
lowest frequency will be = 
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = 
..............2
put here value
wavelength = 
wavelength = 0.08575 m
so distance = 
..............3
distance = 
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Calculate the change in heat of the aluminum; show all calculations. Calculate the change in heat of the water; show all calculations. Are the two values the same? Why or why not? See the attached picture for the numbers.
I got -3443.14 J for the aluminum and 3443.595 for the water
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by
where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is
And the frequency of the motion instead is given by:
(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:
where
F is the force applied initially to the spring, so it is equal to the weight of the block:
k = 160 N/m is the spring constant
Solving for A, we find
Electrical Energy, because it can be transformed into many forms.
